Borrar filtros
Borrar filtros

Discrete transfer fcn 1/z-1 (Z-transform) How it Works?

8 visualizaciones (últimos 30 días)
Hi,
i would like to ask you for little explanation of a simple model.
In this model is one Constant block with value = 1.
The second block is discrete transfer fcn with Z-Transform in this form : 1/z-1
time is set to 2 sec and the sample time of trans. fcn blok to 1. the result is 2 but why? how it works.. this concrete function 1/z-1
and in my view it works like this: 1*1^(-1) + 1*^(-2) = 2 (but im not sure, plz answer if its wrong or correct)
my next question is, can be z=1 , i think "z" in form at the page i pasted? or what the "z" is?
thank you very much. i hope its easy but i realy wondering about it a lot.

Respuesta aceptada

Azzi Abdelmalek
Azzi Abdelmalek el 25 de Mayo de 2013
Editada: Azzi Abdelmalek el 25 de Mayo de 2013
If the input of your system is u[n] and its output is Y[n] then
% H(z)=Y(z)/U(z)
% 1/(z-1)=Y(z)/U(z)
% which gives Y(z)(z-1)=U(z)
% zY(z)-Y(z)=U(z)
% In the temporal domain
% y[n+1]-y[n]=u[n]
% y[n+1]=y[n]+u[n]
% finaly: y[n]=y[n-1]+u[n-1]
% In your example
% y[1]=y[0]+u[0]= 0+1
% y[2]=y[1]+u[1]=1+1=2
% y[3]=y[2]+u[2]=2+1=3
% and so on
  1 comentario
Zsolt
Zsolt el 25 de Mayo de 2013
Editada: Zsolt el 25 de Mayo de 2013
Thank you for fast answer. And can u tell me please if this is correct? http://i.imgur.com/GH6MlRZ.png or how can i get the same result with this form of sum?

Iniciar sesión para comentar.

Más respuestas (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by